$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{ab + bc + ca}{a^{2} + b^{2} + c^{2}} ≤ \frac{5}{2}$
$⇔ \frac{1}{2} + \frac{ab + bc + ca}{a^{2} + b^{2} + c^{2}} ≤ 1 - \frac{a}{b + c} + 1 - \frac{b}{c + a} + 1 - \frac{c}{a + b}$
$⇔ \frac{(a + b + c)^{2}}{2(a^{2} + b^{2} + c^{2})} ≤ \frac{b + c - a}{b + c} + \frac{a + c - b}{a + c} + \frac{b + a - c}{b + a} = P$
+ Đặt:
$x = b + c - a$; $y = a + c - b$; $z = b + a + c$; $x, y, z > 0$.
$a + b = \frac{x + y + 2z}{2}$; $b + c = \frac{y + z + 2x}{2}$; $a + c = \frac{x + z + 2y}{2}$.
$a = \frac{x + z}{2}$; $b = \frac{x + z}{2}$; $c = \frac{x + y}{2}$.
$⇒2(a^{2} + b^{2} + c^{2}) = 2.\frac{y^{2} + 2yz + z^{2} + x^{2} + 2xz + z^{2} + x^{2} + 2xy + y^{2}}{4} = x^{2} + y^{2} + z^{2} + xy + z + zx$
$P = \frac{b + c - a }{b + c} + \frac{a + c - b}{a + c} + \frac{b + a - c}{b + a}$
$⇒P = \frac{2x}{2x + y + z} + \frac{2y}{x + 2y + z} + \frac{2z}{x + y + 2z}$
$= \frac{2x^{2}}{2x^{2} + y + z} + \frac{2y^{2}}{x + 2y^{2} + z} + \frac{2z^{2}}{x + y + 2z^{2}}$
$P ≥ \frac{2(x + y + z)^{2}}{2(x^{2} + y^{2} + z^{2} + xy + yz + zx)}$
$= \frac{(a + b + c)^{2}}{x^{2} + y^{2} + z^{2} + xy + yz + zx } = \frac{(a + b + c)^{2}}{2(a + b + c)^{2}}$
+ Hay: $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} + \frac{ab + bc + ca}{a^{2} + b^{2} + c^{2}} ≤ \frac{5}{2}$
+ Dấu $=$ xảy ra $⇔ a = b = c$ (đpcm).
+ Hay: tam giác đã cho là tam giác đều.
CHÚC EM HỌC TỐT. XIN HAY NHẤT.
